exam1solution_pdf - Version 444 Exam 1 Sutcliffe (52440) 1...

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Unformatted text preview: Version 444 Exam 1 Sutcliffe (52440) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points For gases that do not react chemically with water, the solubility of the gas in water gen- erally (decreases, increases) with an increase in the pressure of the gas and (decreases, in- creases) with increasing temperature. 1. decreases; decreases 2. decreases; increases 3. increases; decreases correct 4. increases; increases Explanation: An increase in pressure means that you have increased the concentration of gas above the solvent surface, thereby increasing the concentration of the gas in the solvent. In- creasing the temperature will decrease the solubility of the gas. 002 10.0 points Calculate the molarity of a solution prepared by dissolving 11 . 5 g NaOH in enough water to make 1 . 5 L of solution. 1. 0.43 M 2. 2.3 M 3. 7.6 M 4. 0.19 M correct Explanation: m NaOH = 11 . 5 g V = 1 . 5 L Molarity is moles solute per liter of solution. We use the molar mass of NaOH (40 g/mol) to convert from grams to moles NaOH: mol NaOH = 11 . 5 g NaOH 1 mol NaOH 40 g NaOH = 0 . 2875 mol NaOH We divide the moles of NaOH in the solution by the total volume of the solution: M = . 2875 mol NaOH 1 . 5 L solution = 0 . 19 M NaOH 003 10.0 points Note: The question is referring to finding the equilibrium constant for the forward reaction. The standard molar Gibbs free energy of formation of NO 2 (g) at 298 K is 51.30 kJ mol 1 and that of N 2 O 4 (g) is 97.82 kJ mol 1 . What is the equilibrium constant at 25 C for the reaction 2 NO 2 (g) N 2 O 4 (g) ? 1. 6.88 correct 2. 9 . 72 10 9 3. None of these 4. 7 . 01 10 9 5. 1.00 6. 0.657 7. 0.145 8. 1 . 02 10 10 Explanation: G products = 97 . 82 kJ mol 1 G reactants = 51 . 30 kJ mol 1 G rxn = summationdisplay n G products- summationdisplay n G reactants = 97 . 82- (2)(51 . 30) = (- 4 . 78 kJ / mol) parenleftbigg 1000 J kJ parenrightbigg =- 4780 J / mol G =- RT ln K K = e G / ( R T ) = exp bracketleftbigg-- 4780 J / mol (8 . 3145 J / mol K)(298 K) bracketrightbigg = 6 . 88395 Version 444 Exam 1 Sutcliffe (52440) 2 004 10.0 points The vapor pressures of pure carbon disulfide and carbon tetrachloride are 360 and 99.8 torr, respectively, at 296 K. What is the vapor pressure of a solution containing 50.0 g of each compound? 1. 460 torr 2. 33.0 torr 3. 241 torr 4. 260 torr 5. 274 torr correct Explanation: 005 10.0 points A certain reaction has H equal to 11.57 kJ/mol. This reaction is normally run at room temperature (25 C). At what new tem- perature should the reaction be run so that K is twice its value at 25 C?...
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exam1solution_pdf - Version 444 Exam 1 Sutcliffe (52440) 1...

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